Question

the peremeter of a triangle is (x²y+10)unit.one of the side of length (x²y-4)unit and another sideis(3-2x²y)unit.Find the length of the third side.

Answer 1

Given : perimeter (P) =(x²y+10)unit

            side one (a)  =(x²y-4)unit

            side two (b)  =(3-2x²y)unit

therefore, To Find : side three (c) = ?

Analysis : Perimeter of Triangle (P) = a+b+c

therefore, $x^{2} y+10=[x^{2} y-4]+[3-2x^{2} y]+c$

$x^{2} +10=-x^{2} y-1+c\\x^{2}y+x^{2}y=c-11\\2x^{2}y=c-11\\c=2x^{2}y+11$

therefore the third side (c) =2x^{2}y+11[/tex]